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Ring Area From A Chord

by profhankd, published

Ring Area From A Chord by profhankd Jun 27, 2014
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Summary

I'm not sure where this puzzle originated; it's old. Arguably, it is not even a puzzle in the sense that it consists of only a single piece, but it's a great way to annoy your normally-good-at-math friends.

The goal is to determine the area of the ring formed by the difference between two concentric circles given only the length of a chord which is tangent to the inner circle. In this case, the chord is, as clearly marked, 10cm long.

Instructions

This is a trivially easy print in one color. The raised text is then highlighted using paint or, as shown here, permanent marker.
What is the area of the ring?


STOP READING HERE IF YOU WANT TO FIGURE IT OUT YOURSELF.

You could solve this puzzle by doing a bit of trig, but using a little idea from calculus is much easier. The key is the realization that the length of the chord is the only thing that matters. If that was not true, the problem would be unsolvable without additional information. Thus, we can freely change the radii of the circles -- and when the inner radius approaches zero, the trig gets very easy. If the inner circle radius is zero, so is the area of the smaller circle that we need to subtract from the area of the larger circle. Even better, the chord becomes the diameter of the outer circle. Thus, we can use the good old Pi(rr) formula. Here, r=10/2 cm, so the area must be 25Pi cm2.

Incidentally, if you want to make people think this problem is even harder, be sure to call the circular ring the annulus. That ought to get them thinking that they need to remember even more trig. ;-)

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SPOILER
Don`t read this, if you want to solve the problem yourself.
Here is another way to solve this:
Think
of a square of width a=10cm. Inscribe a circle that touches the inside
of the square and "out-scribe" a circle that touches all corners. The
answer will be the same as in the instruction.

** SPOILER ****
Don't try to bend the circle. Instead realize that there is no inner circle.

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